# Show that the dimension of the set of all linear transformations from $V$ to $W$ has dimension equal to $mn$.

Let $V, W$ be vector spaces. The set $L(V,W)$ is the set of all linear transformations $T: V \to W$, with the operations:

$(T_1 + T_2)(v) = T_1(v)+T_2(v)$

$(\alpha T)(v) = \alpha T(v)$

for $v\in V$.

Prove that the dimension of the set $L(V,W)$ is $\mathrm{dim}(L(V,W)) = \mathrm{dim}(V) \cdot \mathrm{dim}(W)$

I know I have to (or can) use the bases of the vector spaces to prove this but I cannot come up with a solution.

• If $V,W$ are finite dimensional: after fixing bases of $V,W$, $T$ corresponds with a matrix having dimension $\text{dim}(W) \times \text{dim}(V)$. Do you know a familiar space to which the space of such matrices is isomorphic? Apr 3, 2017 at 9:49
• The vector space $L(V,W)$? My problem is that I don't know how to argue that. Apr 3, 2017 at 9:55
• also. But it is isomorphic to another familiar space: consider a matrix, then we can make it into a rowvector by pasting row to row, hence we find some element of $\mathbb{R}^{mn}$. What is its dimension? Apr 3, 2017 at 9:57
• Note that each linear transformation corresponds with exactly one matrix if we fix a basis for $V$ and $W$. Hence $L(V,W)$ is isomorphic to the set of matrices I described. Apr 3, 2017 at 10:02
• Can yoiu assume that the dimensions are finite? Otherwise, note that if $\dim V=\aleph_0$ and $\dim W=1$, we have $\dim L(V,W)=2^{\aleph_0}$. Apr 3, 2017 at 10:06

A longer hint:

Pick a basis $\{e_i\}$ of V and a basis $\{f_j\}$ of W. Consider the set of linear transformations $S=\{T_{mn}\}$ where we define $T_{mn}$ as $$T_{mn}(\sum a_ie_i) =a_mf_n$$ In other words, $T_{mn}$ maps $e_m$ to $f_n$ and maps the other basis vectors of V to 0 (in W), and then we extend this over all of V while keeping $T_{mn}$ linear.

Can you show that the transformations in S are independent i.e. you cannot make one of them by adding together multiples of the others ?

Can you see how to add together multiples of transformations in S to create a transformation that maps $e_m$ in V to any given vector w in W ?

Now note that a linear transformation in T is defined by its action on each of the basis vectors in V. Can you see how to add together multiples of transformations in S to create any given linear transformation in T ?

Once you have shown that the transformations in S are independent and that they cover all of T, then you know that S is a basis of T. Count the size of S to find the dimension of T

Hint:

By definition of a basis, if $\mathcal B$ is a basis of $V$, and we set $m=\dim V$, $$\mathcal L(V,W)\simeq W^{\mathcal B}\simeq W^m,\enspace\text{so}\quad \dim\mathcal L(V,W)=\dim W^m=m\dim W.$$

This explanation helps if you know that two finite-dimensional vector spaces over the same field, $$F$$, are isomorphic if and only if they have equal dimensions.

If you're using Sheldon Axler's Linear Algebra Done Right (3rd. Edition), you might have seen $$\mathcal{L}(V,W)$$ as a domain in the context of using the linear map $$\mathcal{M}$$, which maps $$\mathcal{L}(V,W)$$ to $$F^{mn}$$, where $$F^{mn}$$ is the set of all m-by-n matrices, where each entry is in $$F$$.

Formally:

$$\mathcal{M} : \mathcal{L}(V,W) \rightarrow F^{mn}$$

So using the Fundamental Theorem of Linear Maps, we know that:

dim $$\mathcal{L}(V,W) =$$ dim null $$\mathcal{M}$$ $$+$$ dim range $$\mathcal{M}$$

So our goal now is to prove that: dim $$\mathcal{L}(V,W) =$$ (dim $$V$$)(dim $$W$$)

Let's start with a basis $$v_1, ..., v_n$$ for $$V$$ and a basis $$w_1, ..., w_m$$ for a basis of W. This means that dim $$V = n$$ and dim $$W = m$$. And this also makes sense because $$\mathcal{M}(T)$$, where $$T \in \mathcal{L}(V,W)$$, represents an m-by-n matrix that holds the coefficients for the basis of $$W$$ to make $$Tv_k$$:

$$Tv_k = A_{1,k}w_1 + A_{2,k}w_2 + ... + A_{m,k}w_m$$, where $$A_{j,k}$$ are entires of the k-th column of $$\mathcal{M}(T)$$ for $$1 \leq j \leq m$$

If you need help convincing yourself that dim $$F^{mn} = mn =$$ (dim $$V$$)(dim $$W$$), just imagine the basis of $$F^{mn}$$ to be a list of m-by-n matrices where each matrix has entries of all $$0$$'s except one where there is a $$1$$:

$$\begin{equation} \begin{bmatrix} 1 & 0 & \ldots & 0\\ 0 & 0 & \ldots & 0\\ \vdots & \ldots & \ldots & 0\\ 0 & \ldots & \ldots & 0\\ \end{bmatrix}, \ \begin{bmatrix} 0 & 1 & \ldots & 0\\ 0 & 0 & \ldots & 0\\ \vdots & \ldots & \ldots & 0\\ 0 & \ldots & \ldots & 0\\ \end{bmatrix}, \ \ldots , \ \ \begin{bmatrix} 0 & 0 & \ldots & 0\\ 0 & 0 & \ldots & 0\\ \vdots & \ldots & \ldots & 0\\ 0 & \ldots & \ldots & 1\\ \end{bmatrix} \end{equation}$$

We see that there are a total of $$mn$$ matrices in the basis of $$F^{mn}$$.

Now, in order for dim $$\mathcal{L}(V,W) = mn =$$ (dim $$V$$)(dim $$W$$), we need to show that dim $$\mathcal{L}(V,W) =$$ dim $$F^{mn}$$. So that means that we need to show that dim null $$\mathcal{M} = 0$$ and dim range $$\mathcal{M} =$$ dim $$F^{mn}$$ (in other words, show that $$\mathcal{M}$$ is injective and surjective).

To show that $$\mathcal{M}$$ is injective, we need to show that $$\mathcal{M}(T) = 0$$, where $$T = 0$$. In order to do so, let's suppose that $$\mathcal{M}(T) = 0$$, where $$0$$ is the m-by-n matrix that has all entries as $$0$$. What this means is that: $$Tv_k = 0$$ because:

$$Tv_k = 0w_1 + 0w_2 + ... + 0w_m$$

Since $$v_k$$ is an element from the basis of $$V$$, $$v_k \neq{0}$$, therefore $$T = 0$$. Hence, null $$\mathcal{M} = \{0\}$$ and thus dim null $$\mathcal{M} = 0$$.

So what we have now is dim $$\mathcal{L}(V,W) =$$ dim range $$\mathcal{M}$$, where we need to show that range $$\mathcal{M} = F^{mn}$$.

Let's suppose that $$A \in F^{mn}$$. Remember that:

$$Tv_k = A_{1,k}w_1 + A_{2,k}w_2 + ... + A_{m,k}w_m$$, where $$A_{j,k}$$ are entires of the k-th column of $$\mathcal{M}(T)$$ for $$1 \leq j \leq m$$

So, we see that $$\mathcal{M}(T) = A$$. By using the element-method, we see that range $$\mathcal{M} \subseteq F^{mn}$$ and $$F^{mn} \subseteq$$ range $$\mathcal{M}$$, thus range $$\mathcal{M} = F^{mn}$$.

Therefore, dim $$\mathcal{L}(V,W) =$$ dim $$F^{mn} = mn =$$ (dim $$V$$)(dim $$W$$).

As we see, we showed that because these two vector spaces are isomorphic, they possess equal dimensions.